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I am trying to tackle the following question, but I actually don't know how to start...

Let $f:\Bbb{R}\to\Bbb{R}$ be measurable function with period $1$ and $\displaystyle \int_{0}^{1}|f|<\infty$.

Define $\displaystyle f_n(x)=f\left(x-\frac{1}{n}\right)$. Show that $$\lim_{n\to\infty}\int\limits_{[0,1]}|f-f_n|=0$$

I have tried to use periodicity of $f$ and manipulate $f_n(x)$, but to no avail.

Please help, thanks!

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1 Answer 1

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Let $L^1(T)$ denote the space of $1$-periodic measurable functions $f$ such that $\int_0^1 \lvert f\rvert < \infty$. Trigonometric polynomials of the form $\sum\limits_{\lvert k \rvert \le N} c_k e^{2\pi i kx}$ are dense in $L^1(T)$, so first prove the result when $f$ is a trigonometric polynomial, then use density.

If $f(x) = \sum\limits_{\lvert k \rvert \le N} c_k e^{2\pi i kx}$, then

$$f(x) - f\left(x - \frac1n\right) = \sum_{\lvert k\rvert \le N} c_ke^{2\pi ix}\left(1 - e^{-2\pi i/n}\right)\tag{*}$$

Using the fact that $\lvert 1 - e^{i\theta}\rvert \le \lvert \theta\rvert$ for all $\theta \in \Bbb R$, we have $\lvert 1 - e^{2\pi i /n}\rvert \le 1/n$. So the sum in (*) is bounded by $C/n$, where $C = \sum\limits_{\lvert k \rvert \le N} \lvert c_k\rvert$. Hence $\|f - f_n\|_{L^1(T)} \le C/n\to 0$ as $n\to \infty$, and the result holds when $f$ is a trigonometric polynomial.

Now let $f\in L^1(T)$ and $\epsilon > 0$. There is trigonometric polynomial $g$ such that $\|f - g\|_{L^1(T)} < \epsilon/3$. By periodicity of $f$ and $g$, $\|f_n - g_n\|_{L^1(T)} = \|f - g\|_{L^1(T)}$. Indeed,

\begin{align}\|f_n - g_n\|_{L^1(T)} &= \int_0^1 \lvert f_n(x) - g_n(x)\rvert\, dx = \int_{-1/n}^{1 - 1/n} \lvert f(x) - g(x)\rvert\, dx\\ &=\int_{-1/n}^0 \lvert f(x) - g(x)\rvert\, dx + \int_0^{1-1/n} \lvert f(x) - g(x)\rvert\, dx\\ &=\int_{1-1/n}^1 \lvert f(x) - g(x)\rvert\, dx + \int_0^{1-1/n} \lvert f(x) - g(x)\rvert\, dx\\ &=\|f-g\|_{L^1(T)}\end{align}

where periodicity was used on first integral in the second-to-last line. Let $n_0$ be a positive integer such that $\|g - g_n\|_{L^1(T)} < \epsilon/3$ whenever $n > n_0$. For all $n > n_0$, \begin{align}\|f - f_n\|_{L^1(T)} &\le \|f - g\|_{L^1(T)} + \|g - g_n\|_{L^1(T)} + \|g_n - f_n\|_{L^1(T)}\\ & = 2\|f - g\|_{L^1(T)} + \|g - g_n\|_{L^1(T)} \\ &< 2\cdot \frac{\epsilon}{3} + \frac{\epsilon}{3} = \epsilon\end{align}

Since $\epsilon$ was arbitrary, the result follows.

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  • $\begingroup$ This is the standard approach, but step functions are also dense in $L^1$. So it suffices in the end to consider the characteristic function of a bounded interval. I wonder some times why text books don't go this route. $\endgroup$ Aug 30, 2016 at 15:04
  • $\begingroup$ Like you said @Harald, step functions are dense in $L^1[0,1]$, but are they $1$-periodic? For instance, $1_{[1,2]}(x+1) = 1_{[0,1]}(x) \neq 1_{[1,2]}(x)$. $\endgroup$
    – kobe
    Aug 30, 2016 at 15:22
  • $\begingroup$ @kobe, thanks for your answer. I still don't understand why $g$ is $1$-periodic. Could you please elaborate? $\endgroup$
    – Galc127
    Aug 30, 2016 at 15:36
  • $\begingroup$ Hi @Galc127, since $e^{2\pi ik} = 1$ for every integer $k$, then $e^{2\pi ik(x+1)} = e^{2\pi ikx}e^{2\pi ik} = e^{2\pi ikx}$ for all $x\in \Bbb R$. Hence $e^{2\pi ikx}$ is a $1$-periodic function of $x$. Therefore, a linear combination of $e^{2\pi ikx}$ is $1$-periodic. $\endgroup$
    – kobe
    Aug 30, 2016 at 15:38
  • $\begingroup$ @kobe, thanks. It seems that I misread your definition of $g$ as a trigonometric polynomial of the form $\sum_{k\le |N|}c_k e^{2\pi i k x}$. $\endgroup$
    – Galc127
    Aug 30, 2016 at 15:40

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